今年是2014年编写程序判断今年是闰年还是平年。
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为什么不能对num初始化 int num = 0;
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#include<stdio.h>
int main(void)
{
int year =2014;
if()
{
printf(")
{
else
{
printf(")
}
C语言中的三目运算符：“?:”，其格式为：
** 表达式1 ? 表达式2 : 表达式3;**
检查条件
这个程序让用户输入一个1~10之间的数字，再确定该数字有多大。
//program 3.1 A simple example of the if statement
#include<stdio.h>
int main (void)'
{
int number = 0 ; //初始化为0，分号为英文下的
printf("\n Enter an integer between 1 and 10: ");
scanf("%d",&number);
if( number > 5 )
printf(" You entered %d which is greater than 5\n", number);
if (number < 6)
printf("You entered %d which is less than 6\n", number);
return 0 ;
}
使用IF语句分析数字
假定某个产品的售价是$3.50/个，当订购数量大于10时，就提供5%的折扣，使用 if-else语句可以计算并输出给定数量的总价。*
//program 3.2 Using if statements to decide on a discount
#include <stdio.h>
int main ()
{
const double unit_price = 3.50;
int quantity = 0;
printf("Enter the number that you want to buy :");
scanf(" %d ", &quantity);
double total = 0.0 ;
if (quantity >10 )
total =quantity*unit_price*0.95;
else
total = quantity*unit_price;
printf("The price for %d is $%.2f\n",quantity,total);
return 0 ;
}
分析数字
下面用另外几个例子练习if技巧。这个程序测试输入的数是偶数还是奇数如果是偶数，就接着测试该数字的一半是否还是偶数*
//program 3.3 Using nested ifs to analyze numbers
#include <stdio.h>
#include <limits.h>
int main (void)
{
long test = 0 L;
printf("Enter an integer less than %1d:", LONG_MAX)
scanf(" %1d", &test );
if (test % 2L == 0L)
{
printf("The number %1d is even", test);
if((test/2l) % 2L== 0L);
{
printf("\nHalf of %1d is also even", test);
printf("\nThat's interesting isn't it?\n")；
}
}
else
printf("The number %1d is odd\n", test);
return 0 ;
}
将大写字母转化为小写字母
这个例子使用新的逻辑运算符，将输入的大写字母转化为小写字母
//program 3.4 converting uppercase to lowercase
#include <stdio.h>
int main (void)
{
char letter = 0;
printf (”Enter an uppercase letter: ");
scanf("%c", &letter);
if (letter >= 'A ')( letter <= 'Z')
{
letter = letter - 'A' + 'a';
printf("You entered an uppercase %c\n", letter);
}
else
printf("Try using the shift key! I want a capital letter.\n");
return 0;
}
转换字母的一种更好的方式
//program 3.5 Testing letters an easier way
#include <stdio.h>
int main (void)
{
char letter = 0;
printf("Enter an upper case letter:");
scanf(" %c ", &letter);
if ((letter >= 'A')&&(letter <= 'z'))
{
letter += 'a'-'A';
printf("You enter an uppercase %c.\n", letter);
}
esle
printf("You did not enter an uppercase letter.\n");
return 0;
}
**使用条件运算符 **
这个折扣业务可以转换为一个小例子。假定产品的单价仍是￥3.50，但提供三个级别的折扣：数量超过50，折扣为15%；数量超过20，折扣为10%；数量超过10，折扣为5%.
//program 3.6 Multiple discount levels
#include <stdio.h>
int main (void)
{
const double unit_price =3.5;
const double discount1 = 0.05;
const double discount2 = 0.1;
const double discount3 =0.15
double total_price = 0.0;
int quantity = 0;
printf("Enter the number that you want to buy: ");
scanf(" %d ", &quantity);
total_price =quantity*unit_prine*(1.0 -
(quantity > 50 ? discount3 : (
(quantity >20 ? discount2 : (
(quantity > 10 ? discount1 : 0.0))));
printf("The price for %d is $%.2f\n",quantity,total_price);
return 0 ；
}
清楚地使用逻辑运算符
假定程序要为一家大型药厂面试求职者。该程序给满足某些教育条件的求职者提供面试的机会。满足如下条件的求职者会接到面试通知*：
（1）25岁以上，化学专业毕业生，但不是毕业于耶鲁。
（2）耶鲁大学化学专业毕业生。
（3）28岁以下，哈弗大学经济学专业毕业生。
（4）25岁以上，耶鲁大学非化学专业毕业生。
实现该逻辑的程序如下：
//program 3.7 confused recruiting policy
#include < stdio.h >
#include < stdbool.h >
int main ( void )
{
int age = 0;
int college = 0;
int subject = 0;
bool interview = false;
printf("\nWhat college? 1 for harvard, 2 for Yale, 3 for other: ");
scanf("%d", &college);
printf("\nwhat subject? 1 for Chemistry, 2 for economics, 3 for other: ")
scanf("%d", &subject);
printf("\nHow old is the applicant? ")；
scanf("%d", &age);
if(age >25 && subject == 1) && (college == 3 || college ==1 )
interview = true ;
if (college == 2 && subject == 1 ) 逻辑与（且）
interview = true;
if(college == 2 && (subject == 2 || subject == 3) && age > 25 )
interview = true;
if(college == 2 &&(subject == 2 || subject == 3 ) && age > 25)
interview = true;
if(interview)
printf("\n\nGive 'em an interview\n");
else
printf("\n\nReject 'em\n");
return 0;
}
试试看 ：选择幸运数字
这个例子假定，在抽奖活动中有三个幸运的数字，参与者要猜测一个幸运的数字，switch语句会结束这个猜测过程，给出参与者可能赢得奖励.*
//program 3.8 Lucky Lotteries
#include<stdio.h>
int main (void)
{
int choice = 0;
printf("pick a number between 1 and 10 and you may a prize ! ");
scanf("%d", &choice);
if ((choice > 10 ) || (choice < 1 )) 逻辑或
choice = 11
switch(choice)
{
case 7:
printf("Congratulations!\n");
printf("You win the collected works of Amos Gruntfutock.\n");
break;
case 2:
printf(" You win the folding thermometer-pen-watch-umbrell.\n")
break;
case 8:
printf("You win the lifetime supply of aspirin tablets.\n")
break;
case 11:
printf("Try between 1 and 10.You wasted your guess.\n")
break;
default:
printf("Sorry, you lose.\n");
break;
}
return 0;
}
试试看：是或否
//program 3.9 testing cases
#include <stdio.h>
int main(void)
{
int main(void)
char answer =0;
printf(" Enter Y or N: ");
scanf(" %c", &answer);
switch(answer)
{
case 'y':case 'y':
printf("You responded in the affirmative.\n");
break;
case 'n': case 'N':
printf("You responded in the negative.\n");
break;
default:
printf("You did not respond correctly...\n");
break;
}
return 0;
}
switch 语法的使用
#include "stdafx.h"
#include "stdio.h"
#define UP 1
#define DOWN 2
#define LEFT 3
#define RIGHT 4
int _tmain(int argc, _TCHAR* argv[])
{
int dir = UP;
switch (dir){
case UP:
printf("GO up\n");
break;
case DOWN:
printf("GO d");
break;
case LEFT:
printf("Go left\n");
break;
case RIGHT:
printf("Go Right\n");
break;
}
return 0;
}
int main (void)
{
int a = 10;
int b = 8;
if(a > b){
printf(" Max number is a, %d, a");
}
else{
printf("Max number is b, %d,b");
}
return 0;
}
int main(void)
{
int score =0;
if(score >80){
printf("Fine\n");
}else if(score >60){
printf("Ok\n");
}esle(score<60){
printf("Fail\n");
return 0;
}